Ok simple model here but here goes.
Take the angle between the NDS spoke and the vertical to be theta _1.
Take the angle between the NS spoke and the vertical to be theta _2.
The tension in the NDS spoke to be T_1.
The tension in the DS spoke to be T_2.
(Draw it)
Considering lateral components on a wheel with equal number of spokes.
T_1 sin theta_1 = T_2 sin theta_2,
sin theta_1/sin theta_2 = T_2/T_1
=R_1,
the ratio between the tensions.
On a dished wheel theta_2 < theta_1 => sin theta_1 > sin theta_2 => r_1 > 1. Which we already know.
Now consider the case where there are a unequal number of spokes, we shall donate this by a ratio s_r where
s_r = n_1/n_2,
where n_1 = number of spokes on NDS and n_2 = number of spoke on drive side. Again considering the lateral components
s_r T’_1 sin theta_1 = T’_2 sin theta_2
s_r sin theta_1/sin theta_2 = T’_2/ T’_1
=R_2
= s_r R_1.
For the wheel with an unequal number of spokes on either side to be stronger.
|1-s_r R_1| < |1-R_1|
although we know R_1 > 1 we don’t know if s_r R_1 > 1 hence the absolute value is needed.
Squaring
1 – 2s_r R_1 +s_2^2 R_1^2 < 1- 2R_1 + R_1^2
(1-s_r^2)R_1<2(1-s_r)
R_1 < 2(1-s_r)/(1-s_r^2).
In your case
s_r = 8 / 16 = 1 / 2
so for this to be stronger than an equal number of spokes
R_1 < 2(1/2)/(3/4)=4/3
=>sin theta_1 < 4/3 sin theta_2.
So if the depending on the amount of dish, maybe, laterally, but radially might be a different case.